The value of equilibrium constant of the reaction $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2}…

The value of equilibrium constant of the reaction $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2$ is 8.0 The equilibrium constant of the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ will be
  1. $\frac{1}{16}$
  2. $\frac{1}{64}$
  3. 16
  4. $\frac{1}{8}$

Solution

$\begin{gathered}
\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \\
\mathrm{K}_1=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]} \\
\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\
K_2=\frac{\left[\mathrm{HI}^2\right.}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}
\end{gathered}$
From Eqs (i) and (ii)
$\begin{array}{l}
& K_1^2=\frac{1}{K_2} \\
\because & K_1=8.0 \\
\therefore & K_2=\frac{1}{K_1^2}=\frac{1}{8^2}=\frac{1}{64}
\end{array}$

Asked in: NEET 2008 (Screening)

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